(D) The period $T$ of a simple harmonic motion is given by $T = 2\pi \sqrt{\frac{m}{k}}$,which shows that the period is independent of the amplitude $

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its period will not change but total energy will decrease

(D) The period $T$ of a simple harmonic motion is given by $T = 2\pi \sqrt{\frac{m}{k}}$,which shows that the period is independent of the amplitude $A$.Therefore,the period will not change.The total energy $E$ of a particle in $S.H.M.$ is given by $E = \frac{1}{2} m \omega^2 A^2$.Since $E \propto A^2$,if the amplitude $A$ is decreased,the total energy $E$ will also decrease.Thus,the period remains constant while the total energy decreases.

Class 11 Physics Oscillations Energy of Simple Harmonic Motion

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If the amplitude of linear $S.H.M.$ is decreased,then:

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A
its period and total energy will increase
B
its period will increase,and total energy will decrease
C
its period and total energy will decrease
D
its period will not change but total energy will decrease

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If the amplitude of linear $S.H.M.$ is decreased,then:

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Starting from the mean position,a body oscillates simple harmonically with a period $T$. After what time will its kinetic energy be $75 \%$ of the total energy? $(\sin 30^{\circ} = 0.5)$

$A$ particle of mass $10 \ g$ performs simple harmonic motion with an amplitude of $10 \ cm$ and a time period of $2 \ s$. What is its kinetic energy at a displacement of $5 \ cm$?

In $SHM$,the restoring force is $F = -kx$,where $k$ is the force constant,$x$ is the displacement,and $A$ is the amplitude of motion. Then,the total energy depends upon:

$A$ body executes simple harmonic motion with an amplitude $A$. At what displacement,from the mean position,is the potential energy of the body one fourth of its total energy?

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